Answer:
(a) See attachment
(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.
(c) 
Explanation:
(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:
For [110]

For [220]

The drawn of the planes is shown in the attachments.
(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:
For (1,1):

For (0.5, 0.5):

As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.
(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:
The Miller indices are already given in the statement. Then, the distance is:


Answer:
Option D
160 kHz
Explanation:
Since we must use at least one synchronization bit, total message signal is 15+1=16
The minimum sampling frequency, fs=2fm=2(5)=10 kHz
Bandwith, BW required is given by
BW=Nfs=16(10)=160 kHz
Answer:
Vb.Net
msgbox ("Press "q" twice to quit", msgboxstyle.information)
if char.q = keypress and keypress.count = 2 then
End
End if
Explanation:
You are going to have to be more specific than that, perhaps add a picture next time ? Unfortunately, this question can’t be answered.
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W